I'd like to know if I understand correctly why $y[n] = x[n-1]$ is an all pass filter.

By definition of the filter, we get the output the same of the input only delayed by one unit of time. Therefore, all frequencies that entered must pass (eventually).

Therefore, it's an all-pass filter.

Is that a correct reasoning?


  • $\begingroup$ would you please fix your notation, so i don't have to? $\endgroup$ – robert bristow-johnson Mar 12 '18 at 21:16

Your notation is a bit odd, as you are mixing the time-domain with the $z$-domain in the equation you wrote. If you refer to the system

$$y[n]=x[n-1] \quad \mathrm{or} \quad Y(z)=z^{-1}X(z)$$

then you are correct. To put it mathematically, the transfer function of the filter would be:


The frequency response is then

$$H(z) \Bigg|_{z=e^{j\omega}} =\frac{1}{e^{j\omega}}$$

You can see that the magnitude of $H(e^{j\omega})$ is $1$ for all frequencies. Therefore, it is an all-pass system.

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  • $\begingroup$ yeah, I meant in your notation $y(n) = x(n-1)$ . Can you please explain how did you reach the transfer function $H(z) = \frac{1}{z}$? $\endgroup$ – deficiencyOn Mar 12 '18 at 17:50
  • $\begingroup$ @deficiencyOn $$Y(z)=z^{-1}X(z) \implies \frac{Y(z)}{X(z)}=z^{-1}=\frac{1}{z}$$ $\endgroup$ – Tendero Mar 12 '18 at 17:51
  • $\begingroup$ @deficiencyOn If you find an answer helpful and clears out the doubt for you, please accept it by clicking on the check sign next to it, so that the post is considered solved and can help future readers. $\endgroup$ – Tendero Mar 14 '18 at 20:19

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